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

ID:(2050, 0)

Iframe

>Top

ID:(15199, 0)

Concept

>Top

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].

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

ID:(2078, 0)

Concept

>Top

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:

ID:(15096, 0)

Description

>Top

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:

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:

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:

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.

ID:(15099, 0)

Top

>Top

Parameters

$\rho_c$

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

Variables

$q_c$

q_c

Clay own porosity

-

$p_c$

p_c

Clay own volume factor

-

$f_c$

f_c

Force per grain

N

$V_m$

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

-

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

Symbol

Equation

Solved

Translated

Calculations

Symbol

Equation

Solved

Translated

Variable Given Calculate Target : Equation To be used

Equations

ID:(15218, 0)

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 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 ($V_t$).

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

Conditions

Variables

$V_a$

Solid volume of sand

$m^3$

6554

$V_t$

Total volume

$m^3$

4946

$f_a$

Volume fraction of sand in the sample

$-$

10105

Relation

ID:(10369, 0)

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 ($V_t$).

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

Conditions

Variables

$V_i$

Solid volume of silt

$m^3$

6555

$V_t$

Total volume

$m^3$

4946

$f_i$

Volume fraction of silt in the sample

$-$

10106

Relation

ID:(10367, 0)

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.

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

Conditions

Variables

$f_c$

Force per grain

$N$

8270

$V_c$

Solid volume of clay

$m^3$

6556

$V_t$

Total volume

$m^3$

4946

Relation

ID:(10368, 0)

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 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 ($V_t$).

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

Conditions

Variables

$V_m$

Macropore volume

$m^3$

10109

$V_t$

Total volume

$m^3$

4946

$f_m$

Volume fraction of macropores in the sample

$-$

10110

Relation

ID:(15084, 0)

Equation

>Top, >Model

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:

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:

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

Conditions

Variables

$q_c$

Clay own porosity

$-$

10104

$q_a$

Sand's own porosity

$-$

10102

$q_i$

Silt own porosity

$-$

10103

$f_c$

Volume fraction of clay in the sample

$-$

10107

$f_m$

Volume fraction of macropores in the sample

$-$

10110

$f_a$

Volume fraction of sand in the sample

$-$

10105

$f_i$

Volume fraction of silt in the sample

$-$

10106

Relation

Development

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)

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 ($V_c$):

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:

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

Conditions

Variables

$q_c$

Clay own porosity

$-$

10104

$f$

Porosity

$-$

5805

$q_a$

Sand's own porosity

$-$

10102

$q_i$

Silt own porosity

$-$

10103

$f_c$

Volume fraction of clay in the sample

$-$

10107

$f_a$

Volume fraction of sand in the sample

$-$

10105

$f_i$

Volume fraction of silt in the sample

$-$

10106

Relation

Development

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)

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 ($V_t$):

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:

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

Conditions

Variables

$\rho_a$

Density of a grain of sand

2.63e+3

$kg/m^3$

10095

$g_a$

Mass fraction of sand in the sample

$-$

5797

$f$

Porosity

$-$

5805

$\rho_s$

Solid Density

$kg/m^3$

4944

$f_a$

Volume fraction of sand in the sample

$-$

10105

Relation

Development

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)

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 ($V_t$):

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:

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

Conditions

Variables

$\rho_i$

Density of a grain of silt

2.65e+3

$kg/m^3$

10094

$g_i$

Mass fraction of silt in the sample

$-$

10098

$f$

Porosity

$-$

5805

$\rho_s$

Solid Density

$kg/m^3$

4944

$f_i$

Volume fraction of silt in the sample

$-$

10106

Relation

Development

ID:(15094, 0)

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 ($V_t$):

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:

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

Conditions

Variables

$\rho_c$

Density of a grain of clay

2.8e+3

$kg/m^3$

10093

$g_c$

Mass fraction of clay in the sample

$-$

10099

$f$

Porosity

$-$

5805

$\rho_s$

Solid Density

$kg/m^3$

4944

$f_c$

Volume fraction of clay in the sample

$-$

10107

Relation

ID:(15095, 0)

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 ($q_c$):

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:

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

Conditions

Variables

$q_c$

Clay own porosity

$-$

10104

$g_c$

Mass fraction of clay in the sample

$-$

10099

$g_a$

Mass fraction of sand in the sample

$-$

5797

$g_i$

Mass fraction of silt in the sample

$-$

10098

$f$

Porosity

$-$

5805

$q_a$

Sand's own porosity

$-$

10102

$q_i$

Silt own porosity

$-$

10103

Relation

Development

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)

Equation

>Top, >Model

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

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

Conditions

Variables

$p_a$

Own volume factor of sand

$-$

6557

$q_a$

Sand's own porosity

$-$

10102

Relation

ID:(15087, 0)

Equation

>Top, >Model

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

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

Conditions

Variables

$q_i$

Silt own porosity

$-$

10103

$p_i$

Silt own volume factor

$-$

6558

Relation

ID:(15088, 0)

Equation

>Top, >Model

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

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

Conditions

Variables

$q_c$

Clay own porosity

$-$

10104

$p_c$

Clay own volume factor

$-$

6559

Relation

ID:(15098, 0)

Equation

>Top, >Model

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

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

Conditions

Variables

$p_p$

Microporosity own volume factor

$-$

6053

$f$

Porosity

$-$

5805

Relation

ID:(2076, 0)

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 ($g_c$),

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

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

Conditions

Variables

$p_c$

Clay own volume factor

$-$

6559

$g_c$

Mass fraction of clay in the sample

$-$

10099

$g_a$

Mass fraction of sand in the sample

$-$

5797

$g_i$

Mass fraction of silt in the sample

$-$

10098

$p_p$

Microporosity own volume factor

$-$

6053

$p_a$

Own volume factor of sand

$-$

6557

$p_i$

Silt own volume factor

$-$

6558

Relation

Development

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)