Storyboard

The porosity of the soil allows rain or irrigation water to penetrate the soil and reach the napa. Therefore we must study how it can be modeled based on our geometric model as the water moves.

>Model

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Concept

>Top

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Concept

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The flux density ($j_s$) can be expressed in terms of the hydraulic conductivity ($K_s$), in the infinitesimal limit with the column height differential ($dh$) and the distance differential ($dx$), as follows:

This means that the steeper the gradient or the steeper the terrain, the larger the flux density ($j_s$) will be, as illustrated in the graph:

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The graph shows how bars with equal distance differential ($dx$) values have progressively smaller column height differential ($dh$) values, resulting in a decreasing flux density ($j_s$). Since the volume of the liquid is conserved, this can only be possible if there is another flow that compensates for this reduction in flux density ($j_s$). This could be a flow perpendicular to the one shown, for example, if the shorter bars are wider in a direction perpendicular to the graph.

This issue leads to the following:

The height $h$ of the liquid can only be calculated as a result of solving a differential equation, as it must meet the requirement that volume is conserved throughout the entire area where flow occurs.

Additionally, it is important to keep in mind that:

The negative sign reflects the fact that flow always goes from the higher to the lower elevation zone. If the slope is negative, the negative sign results in positive flow (from left to right), and conversely, if the slope is positive, the flow is negative (from right to left).

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Concept

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If we study the one-dimensional case, describing the process along the $x$-axis, we can observe how the height of the column $\Delta h$ varies over a time interval $\Delta t$. In this case, a column with width $\Delta x$ will change its volume per unit length over time as $\Delta x \Delta h/\Delta t$. On the other hand, the amount of liquid entering along the column at $x$ is $h(x) j_s(x)$, while at $x+\Delta x$ it exits as $h(x+\Delta x) j_s(x+\Delta x)$:

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Therefore, the variation of the height of the water column on the ground ($h$) over time is equal to the variation of the product of the height of the water column on the ground ($h$) and the flux density ($j_s$) at position:

Partial derivatives are similar to ordinary derivatives, with the difference that they are applied to functions that depend on more than one variable. In these cases, the partial derivative, denoted by the symbol $\partial$, reminds us of the typical derivative denoted by the letter $d$, but with the peculiarity that the variables not mentioned in the denominator are held constant.

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Concept

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In the case of flow towards a channel, the system can be modeled in a one-dimensional manner, where the height of the water column on the ground ($h$) is a function of the position of the water column on the ground ($x$) representing the flux density ($j_s$), and it satisfies the condition

with the flow at a reference point ($j_{s0}$) and the reference height of the water column ($h_0$) defining the water profile in the soil:

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The key to this equation is that the product of the height of the water column on the ground ($h$) and the flux density ($j_s$) must always remain constant. In this sense, if the height of the water column on the ground ($h$) increases, the flux density ($j_s$) decreases, and vice versa. Moreover, the sign remains the same; hence, flow towards the channel, i.e., negative flow, will occur only when the groundwater level is higher than that of the channel. As the liquid approaches the channel, the groundwater level decreases, leading to an increase in flow density.

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Concept

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The solution to the one-dimensional flow equation towards a channel, where the height of the water column on the ground ($h$) is calculated as a function of the reference height of the water column ($h_0$) and the position of the water column on the ground ($x$) at the channel's edge, along with the characteristic length of the flow in the ground ($s_0$), takes the following form:

This solution is graphically represented in terms of the additional factors $h/h_0$ and $x/s_0$ as follows:

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The profile reveals that, away from the channel, the height of the water column is significantly high. However, due to water extraction by the channel, this height begins to decrease until it reaches the channel's edge. Dynamically, the flux density ($j_s$) determines the amount of water flowing into the channel, while the reference height of the water column ($h_0$) gradually adjusts until it reaches an equilibrium state. In other words, if the value of the reference height of the water column ($h_0$) is too low relative to the total amount of water arriving at the channel, it increases; and if it is too high, it decreases. In this way, the reference height of the water column ($h_0$) acquires the value that balances the amount of incoming water with the amount of water flowing through the channel.

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Concept

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The solution obtained for the height and the parameters the flow at a reference point ($j_{s0}$) and the reference height of the water column ($h_0$) reveals that the flux density ($j_s$) is given by:

We can graphically represent the flux density ($j_s$) as a function of the additional factors $j_s/j_{s0}$ and $x/s_0$ as follows:

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It is noticeable that the flux density ($j_s$) continues to increase as we approach the channel, as the height of the water column on the ground ($h$) decreases. This increase is necessary to maintain the flow velocity in the flux density ($j_s$) or, alternatively, to increase it.

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Concept

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In the case where the flow emerges from the channel, a situation arises where the level of the height of the water column on the ground ($h$) must decrease as we move away from the channel, ensuring the existence of the pressure gradient that drives the flow. The problem is that if the flow rapidly moves within the medium, the height will tend to zero, and as a result, the flow will approach infinity, which doesn't make sense.

This means that there is no stationary solution in such a scenario, and the only solution is for the medium to fill up until it reaches the height of the channel, effectively becoming constant.

The question is whether there exists a non-trivial stationary situation that represents a real and interesting scenario. One possible case is when the level of the medium decreases to the point where it becomes lower than the column before the solution diverges. This case corresponds to the situation where the flow emerges at the surface, and there is no divergence in the solution. This would imply that a flow is generated that exits to the exterior at a certain point, with the risk of weakening the foundation and thereby destabilizing the medium, which acts as a dam.

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Concept

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If we consider a situation where the flow from the channel can emerge at the surface, we have a scenario where the flow enters and then exits the medium, making the solution viable.

Emerging at the surface simply implies that the height of the liquid column becomes higher than that of the surrounding medium. In fact, similar to the case of flow towards a channel, this would generate water on the surface, which, if not allowed to flow, would actually form a new channel.

In the case of flow from a channel, it is possible to model the system in a one-dimensional manner, where the height of the water column on the ground ($h$) is a function of the position of the water column on the ground ($x$) representing the flux density ($j_s$) and satisfying the condition:

With the flow at a reference point ($j_{s0}$) and the reference height of the water column ($h_0$) defining the water profile in the soil, as shown in the following image:

The key to the equation lies in the fact that the product of the height of the water column on the ground ($h$) and the flux density ($j_s$) must remain constant at all times. In this regard, if the height of the water column on the ground ($h$) increases, the flux density ($j_s$) will decrease, and vice versa. Furthermore, the sign remains the same. Therefore, flow from the channel, i.e., positive flow, will occur only if the height of the channel is greater than that of the point where the flow emerges. As the liquid moves away from the channel, the height will decrease, and the flow density will increase.

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Concept

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The solution to the one-dimensional flow equation from a channel, in which the value of the height of the water column on the ground ($h$) is calculated as a function of the reference height of the water column ($h_0$) and the position of the water column on the ground ($x$) at the channel's edge, along with the characteristic length of the flow in the ground ($s_0$), takes the following form:

This solution is graphically represented in terms of the additional factors $h/h_0$ and $x/x_0$ as follows:

The profile reveals that the height decreases as one moves away from the channel to maintain a pressure gradient. However, a problem arises when the distance reaches half of the characteristic length of the flow in the ground ($s_0$), as the height of the column reaches zero, and there is no solution for greater distances (the argument of the square root becomes negative). In other words, for the solution to make sense, there must be a mechanism that removes liquid before reaching this critical distance.

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Concept

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The solution obtained for the height and the parameters the flow at a reference point ($j_{s0}$) and the reference height of the water column ($h_0$) shows us that the flux density ($j_s$) is equal to:

We can graphically represent the flux density ($j_s$) in terms of the additional factors $j_s/j_{s0}$ and $x/x_0$ as follows:

the flux density ($j_s$) continues to increase as we approach the channel, as the height of the water column on the ground ($h$) decreases. This increase is necessary to maintain the flow velocity in the flux density ($j_s$) or, alternatively, to increase it.

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Concept

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An example that illustrates the effect of flow through the base in the case of a dam occurred at Dam 1 of the 'Córrego do Feijão' mine in Brumadinho, Minas Gerais, Brazil.

On January 25, 2019, Dam 1, which is shown in the center of the image, collapsed as depicted in images 1 to 6. Initially, the base began to move while the top started to sink. Eventually, a torrent of water emerged from the base as the entire structure collapsed. In the lower central image, you can see the situation after the dam had completely emptied from the side that contained it ([1], [2]):

The upper-left image shows the dam before the collapse, and the diagram explains how the water pushes against the base surface (blue arrows) and causes the center to collapse (beige arrow). The images show the structure again before the collapse (top-right photo), when the base is being forced, causing the upper part to collapse (bottom-left photo), and the resulting water flow at the base (bottom-right photo) [3]:

The dynamics are driven by the high pressure and flow that exist at the base, explaining the emergence of water through this path.

In this case, there were multiple signs of danger, leading to detailed satellite monitoring of the movement of multiple points for over a year. The points are indicated in the upper photo, and in the lower left image, you can see a detail of the base. Specifically, the points that experienced the most total displacement (Bs and Bp) are highlighted, which are also shown in the graph on the right. The graph also shows the amount of rainfall, which contributes to some extent but is not necessarily a key factor [4]:

This example aims to demonstrate how high pressure at the base, combined with a high water flow, contributes to the observed dynamics, without necessarily explaining when or how it became unstable. This will be explored further.

[1] Google Earth Pro for Brumadinho, Minas Gerais, Brazil, January 2019 and February 2019

[2] Vale S.A. cameras

[3] Criminal Investigative Procedure No. MPMG-0090.19.000013-4, Police Investigation No. PCMG-7977979, MINISTÉRIO PÚBLICO DO ESTADO DE MINAS GERAIS

[4] Deformations Prior to the Brumadinho Dam Collapse Revealed by Sentinel-1 InSAR Data Using SBAS and PSI Techniques, Fábio F. Gama, José C. Mura, Waldir R. Paradella, and Cleber G. de Oliveira, MDPI, Remote Sens. 2020, 12, 3664.

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Concept

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In the case of groundwater flow towards a well, the height of the water column on the ground ($h$) as a function of the radius from center of well ($r$) with the water well radius ($r_0$), the characteristic length of the flow in the ground ($s_0$), and the reference height of the water column ($h_0$) is represented by

which defines the water profile in the ground:

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Concept

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The solution to the one-dimensional flow equation towards a well, in which the value of the height of the water column on the ground ($h$) is calculated as a function of the radius from center of well ($r$), the reference height of the water column ($h_0$), and the water well radius ($r_0$) at the well's edge, along with the characteristic length of the flow in the ground ($s_0$), takes the following form:

This solution is graphically represented in terms of the additional factors $h/h_0$ and $r/r_0$ for various $r_0/s_0$ as follows:

The profile reveals that, far from the well, the height of the water column is significantly high. However, due to water extraction by the well, this height begins to decrease until it reaches the well's edge. Dynamically, the flux density ($j_s$) determines the amount of water flowing towards the well, while the reference height of the water column ($h_0$) gradually adjusts itself to reach an equilibrium state. In other words, if the value of the reference height of the water column ($h_0$) is too low relative to the total amount of water reaching the well, it increases, and if it is too high, it decreases. In this way, the reference height of the water column ($h_0$) acquires the value that balances the amount of water arriving with the amount of water being extracted through the well.

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Concept

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The solution obtained for the height and parameters the flow at a reference point ($j_{s0}$) and the radius from center of well ($r$), the water well radius ($r_0$), the characteristic length of the flow in the ground ($s_0$) shows us that the flux density ($j_s$) is equal to:

This solution is graphically represented in terms of the additional factors $j_s/j_{s0}$ and $r/r_0$ for various values of $r_0/s_0$ as follows:

None

the flux density ($j_s$) continues to increase as we approach the channel, while the height of the water column on the ground ($h$) decreases. This increase is necessary to maintain the flow velocity in the flux density ($j_s$) or, alternatively, to increase it.

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Concept

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Selected parameter

Calculations

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Equation

>Top, >Model

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

$J_t$

Flujo Total 2 Capas

$m^3/s$

$J_1$

Own volume factor of sand

$-$

$J_2$

Silt own volume factor

$-$

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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Equation

>Top, >Model

The hydraulic conductivity ($K_s$) represents how the liquid behaves within the medium. Part of the hydraulic conductivity ($K_s$) is inherent to the properties of the medium itself, while another part contains constants that describe the behavior of the liquid. Therefore, it makes sense to introduce a new constant that is specific to the medium and not to the flowing liquid.

As a result, the hydraulic conductivity ($k_s$) is related to the radius of a generic grain ($r_0$), the porosity ($f$), and the generic own porosity ($q_0$) through the following definition:

$ k_s \equiv \displaystyle\frac{ r_0 ^2 }{8 q_0 }\displaystyle\frac{ f ^3 }{(1- f )^2}$

Conditions

Variables

$S_1$

Section in point 1

$m^2$

$S_2$

Section in Point 2

$m^2$

$S_3$

Section in point 3

$m^2$

$S$

Section or Area

$m^2$

Relation

Development

Since the hydraulic conductivity ($K_s$) 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 ($\eta$), and the generic own porosity ($q_0$) through

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

we can define the part that depends solely on the soil as the hydraulic conductivity ($k_s$), expressing it as follows:

$ k_s \equiv \displaystyle\frac{ r_0 ^2 }{8 q_0 }\displaystyle\frac{ f ^3 }{(1- f )^2}$

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Equation

>Top, >Model

The hydraulic conductivity ($k_s$) can be calculated from the hydraulic conductivity ($K_s$), the liquid density ($\rho_w$), the gravitational Acceleration ($g$), and the viscosity ($\eta$) using the following expression:

$ k_s = \displaystyle\frac{ \eta }{ \rho_w g } K_s $

Conditions

Variables

Relation

Development

The hydraulic conductivity ($k_s$) is related to the radius of a generic grain ($r_0$), the porosity ($f$), and the generic own porosity ($q_0$), it is equal to

$ k_s \equiv \displaystyle\frac{ r_0 ^2 }{8 q_0 }\displaystyle\frac{ f ^3 }{(1- f )^2}$

Therefore, with the equation for the hydraulic conductivity ($K_s$), along with the liquid density ($\rho_w$), the gravitational Acceleration ($g$), and the viscosity ($\eta$),

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

it results in the relationship between the hydraulic conductivity ($k_s$) and the hydraulic conductivity ($K_s$) as

$ k_s = \displaystyle\frac{ \eta }{ \rho_w g } K_s $

Typically, soil characterization measurements are performed using a specific liquid, resulting in a value of a hydraulic conductivity ($K_s$). With this value, you can calculate the hydraulic conductivity ($k_s$) using the equation mentioned above.

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Equation

>Top, >Model

The flux density ($j_s$) can be expressed in terms of the height of the water column on the ground ($h$) or, based on the water column pressure ($p_t$) generated by the liquid column. Using the definition of the hydraulic conductivity ($k_s$) in terms of the hydraulic conductivity ($K_s$), we obtain the following expression for the viscosity ($\eta$) and the position of the water column on the ground ($x$):

$ j_s = -\displaystyle\frac{ k_s }{ \eta }\displaystyle\frac{ dp }{ dx }$

Conditions

Variables

Relation

Development

the pressure difference ($\Delta p$) in relation to the liquid density ($\rho_w$), the gravitational Acceleration ($g$), and the height difference ($\Delta h$) is calculated as per the following equation:

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

in the infinitesimal limit where the pressure difference ($\Delta p$) equals the pressure differential ($dp$), denoted as:

$\Delta p \rightarrow dp$

and where the height difference ($\Delta h$) equals the column height differential ($dh$), denoted as:

$\Delta h \rightarrow dh$

Using the relationship of the flux density ($j_s$) with the hydraulic conductivity ($K_s$), the column height differential ($dh$), and the distance differential ($dx$), which is expressed as:

$$

and the relationship for the hydraulic conductivity ($k_s$) with the viscosity ($\eta$), which is expressed as:

$ k_s = \displaystyle\frac{ \eta }{ \rho_w g } K_s $

We can derive the following equation:

$ j_s = -\displaystyle\frac{ k_s }{ \eta }\displaystyle\frac{ dp }{ dx }$

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Equation

>Top, >Model

Given that the one-dimensional equation for the flux density ($j_s$) is expressed as the hydraulic conductivity ($K_s$), the height of the water column on the ground ($h$), and the position of the water column on the ground ($x$) as follows:

It is possible to generalize this equation for the case of a homogeneous medium, resulting in an equation for the flow density in more than one dimension ($\vec{j}_s$) where the hydraulic conductivity ($K_s$) remains constant, as follows:

$ \vec{j}_s = - K_s \nabla h $

Conditions

Variables

Relation

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Equation

>Top, >Model

If we generalize the equation in one dimension for the height of the water column on the ground ($h$) as a function of the time ($t$) and the position of the water column on the ground ($x$) with the flux density ($j_s$):

and replace the partial derivative with a divergence, we obtain with the flow density in more than one dimension ($\vec{j}_s$):

$\displaystyle\frac{\partial h}{\partial t} = - \vec{\nabla} \cdot ( h \vec{j}_s )$

Conditions

Variables

Relation

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Equation

>Top, >Model

In the case of the one-dimensional height equation, expressed as:

We can study the stationary case, which implies that the height of the water column on the ground ($h$) divided by the flux density ($j_s$) must be constant and, in particular, can take values at a specific point denoted by the reference height of the water column ($h_0$) and the flow at a reference point ($j_{s0}$):

$ h j_s = h_0 j_{s0} $

Conditions

Variables

Relation

Development

If, for the height of the water column on the ground ($h$) divided by the flux density ($j_s$), the equation

$\displaystyle\frac{\partial h}{\partial t} = - \displaystyle\frac{\partial}{\partial x}( h j_s )$

in the stationary case reduces to

$\displaystyle\frac{d}{dx} (h j_s) = 0$

which corresponds to the product of the height of the water column on the ground ($h$) and the flux density ($j_s$) being constant. If you have values for a specific point defined by the reference height of the water column ($h_0$) and the flow at a reference point ($j_{s0}$), then you have:

$ h j_s = h_0 j_{s0} $

Note: The differential equation is an ordinary differential equation because it depends solely on the position $x$ and no longer on time $t$.

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Equation

>Top, >Model

With the hydraulic conductivity ($K_s$), the flow at a reference point ($j_{s0}$), and the reference height of the water column ($h_0$), a characteristic length of the flow in the ground ($s_0$) can be defined as follows:

$ s_0 \equiv \displaystyle\frac{| j_{s0} |}{ K_s h_0 }$

Conditions

Variables

Relation

To avoid complicating the analysis, we have defined the expression while considering the absolute value of the flow at a reference point ($j_{s0}$), thus preventing situations where it might be negative. This means that, depending on the sign of the flow at a reference point ($j_{s0}$), we must express the relationship assuming a positive or negative value for the derivative, thereby defining the flow direction.

ID:(4747, 0)

Equation

>Top, >Model

The differential equation for calculating the height of the water column on the ground ($h$) as a function of the position of the water column on the ground ($x$), the reference height of the water column ($h_0$), and the characteristic length of the flow in the ground ($s_0$) is:

$ h \displaystyle\frac{ dh }{ dx } = \displaystyle\frac{ h_0 ^2 }{ s_0 } $

Conditions

Variables

Relation

Development

The equation for the product of the height of the water column on the ground ($h$) and the flux density ($j_s$) as a function of the reference height of the water column ($h_0$) and the flow at a reference point ($j_{s0}$) is:

$ h j_s = h_0 j_{s0} $

And with the equation that describes the flux density ($j_s$) in terms of the hydraulic conductivity ($K_s$) and the position of the water column on the ground ($x$):

$$

And with the expression for the characteristic length of the flow in the ground ($s_0$):

$ s_0 \equiv \displaystyle\frac{| j_{s0} |}{ K_s h_0 }$

We can derive the resulting equation as follows:

$ h \displaystyle\frac{ dh }{ dx } = \displaystyle\frac{ h_0 ^2 }{ s_0 } $

ID:(15108, 0)

Equation

>Top, >Model

The equation for the height of the water column on the ground ($h$) as a function of the characteristic length of the flow in the ground ($s_0$) and dependent on the reference height of the water column ($h_0$) and the characteristic length of the flow in the ground ($s_0$) is as follows:

This equation can be solved analytically for the given conditions with the reference height of the water column ($h_0$) and the characteristic length of the flow in the ground ($s_0$) at the canal boundary as follows:

$ \displaystyle\frac{ h }{ h_0 } = \sqrt{1 + \displaystyle\frac{ 2 x }{ s_0 }} $

Conditions

Variables

Relation

Development

The equation for the height of the water column on the ground ($h$) as a function of the characteristic length of the flow in the ground ($s_0$) and dependent on the reference height of the water column ($h_0$) and the characteristic length of the flow in the ground ($s_0$) is as follows:

$ h \displaystyle\frac{ dh }{ dx } = \displaystyle\frac{ h_0 ^2 }{ s_0 } $

We can rearrange it to facilitate integration as follows:

$h dh = \displaystyle\frac{h_0^2}{x_0}dx$

Then, integrating with respect to $h_0$, the height at the origin, we get:

$\displaystyle\frac{1}{2}(h^2 - h_0^2) =\displaystyle\frac{h_0^2}{s_0}x$

This leads us to the following expression:

$ \displaystyle\frac{ h }{ h_0 } = \sqrt{1 + \displaystyle\frac{ 2 x }{ s_0 }} $

ID:(15105, 0)

Equation

>Top, >Model

The flux density ($j_s$) is related to the hydraulic conductivity ($K_s$), the height of the water column on the ground ($h$), and the position of the water column on the ground ($x$), resulting in

Therefore, with the solution for the flux density ($j_s$) and the flow at a reference point ($j_{s0}$) given the position of the water column on the ground ($x$) and the characteristic length of the flow in the ground ($s_0$), we obtain:

$ \displaystyle\frac{ j_s }{ j_{s0} } = \displaystyle\frac{1}{\sqrt{1 + \displaystyle\frac{ 2 x }{ s_0 }}} $

Conditions

Variables

Relation

Development

With the solution for the flux density ($j_s$) and the flow at a reference point ($j_{s0}$) given the position of the water column on the ground ($x$) and the characteristic length of the flow in the ground ($s_0$), we have:

$ \displaystyle\frac{ h }{ h_0 } = \sqrt{1 + \displaystyle\frac{ 2 x }{ s_0 }} $

We can calculate the flux density ($j_s$) with the hydraulic conductivity ($K_s$) using:

$ j_s = - K_s \displaystyle\frac{ dh }{ dx }$

and employing

$ s_0 \equiv \displaystyle\frac{| j_{s0} |}{ K_s h_0 }$

this way, we obtain:

$ \displaystyle\frac{ j_s }{ j_{s0} } = \displaystyle\frac{1}{\sqrt{1 + \displaystyle\frac{ 2 x }{ s_0 }}} $

ID:(15106, 0)

Equation

>Top, >Model

The differential equation to calculate the height of the water column on the ground ($h$) in terms of the position of the water column on the ground ($x$), the reference height of the water column ($h_0$), and the characteristic length of the flow in the ground ($s_0$) is as follows:

$ h \displaystyle\frac{ dh }{ dx } = -\displaystyle\frac{ h_0 ^2 }{ x_0 } $

Conditions

Variables

$L$

Capillary length

$m$

$J_V$

Flow

$m^3/s$

$\Delta h$

Height or depth difference

$m$

$K_s$

Hydraulic conductivity

$m/s$

$S_t$

Section Soil

$m^2$

Relation

In this case, the sign of the slope is negative because the height must decrease to generate the necessary pressure gradient for fluid movement.

ID:(4369, 0)

Equation

>Top, >Model

The equation for the height of the water column on the ground ($h$) as a function of the characteristic length of the flow in the ground ($s_0$), which depends on the reference height of the water column ($h_0$) and the characteristic length of the flow in the ground ($s_0$), is as follows:

This equation can be analytically solved for the given conditions with the reference height of the water column ($h_0$) and the characteristic length of the flow in the ground ($s_0$) at the channel's edge as follows:

$ \displaystyle\frac{ h }{ h_0 } = \sqrt{1 - \displaystyle\frac{ 2 x }{ x_0 }} $

Conditions

Variables

Relation

Development

The equation for the height of the water column on the ground ($h$) as a function of the characteristic length of the flow in the ground ($s_0$), which depends on the reference height of the water column ($h_0$) and the characteristic length of the flow in the ground ($s_0$), is as follows:

$ h \displaystyle\frac{ dh }{ dx } = -\displaystyle\frac{ h_0 ^2 }{ x_0 } $

We can rearrange it to facilitate integration as follows:

$h dh = -\displaystyle\frac{h_0^2}{x_0}dx$

Then, integrating with respect to $h_0$, the height at the origin, we obtain:

$\displaystyle\frac{1}{2}(h^2 - h_0^2) =-\displaystyle\frac{h_0^2}{x_0}x$

This leads us to the following expression:

$ \displaystyle\frac{ h }{ h_0 } = \sqrt{1 - \displaystyle\frac{ 2 x }{ x_0 }} $

ID:(2214, 0)

Equation

>Top, >Model

The flux density ($j_s$) is related to the hydraulic conductivity ($K_s$), the height of the water column on the ground ($h$), and the position of the water column on the ground ($x$), resulting in:

Therefore, with the solution for the flux density ($j_s$) and the flow at a reference point ($j_{s0}$) given the position of the water column on the ground ($x$) and the characteristic length of the flow in the ground ($s_0$), we obtain:

$ \displaystyle\frac{ j_s }{ j_{s0} } = \displaystyle\frac{1}{\sqrt{1 - \displaystyle\frac{ 2 x }{ x_0 }}} $

Conditions

Variables

$D$

Soil layer length

$m$

Relation

Development

With the solution for the flux density ($j_s$) and the flow at a reference point ($j_{s0}$) given the position of the water column on the ground ($x$) and the characteristic length of the flow in the ground ($s_0$), we obtain:

$ \displaystyle\frac{ h }{ h_0 } = \sqrt{1 - \displaystyle\frac{ 2 x }{ x_0 }} $

We can calculate the flux density ($j_s$) with the hydraulic conductivity ($K_s$) using:

$ j_s = - K_s \displaystyle\frac{ dh }{ dx }$

and utilizing

$ s_0 \equiv \displaystyle\frac{| j_{s0} |}{ K_s h_0 }$

this way, we obtain:

$ \displaystyle\frac{ j_s }{ j_{s0} } = \displaystyle\frac{1}{\sqrt{1 - \displaystyle\frac{ 2 x }{ x_0 }}} $

ID:(4742, 0)

Equation

>Top, >Model

The equation for the height of the water column on the ground ($h$) with the time ($t$) and the flow density in more than one dimension ($\vec{j}_s$) is:

In the stationary case and using the equation for the flow density in more than one dimension ($\vec{j}_s$), when we expand the derivatives, we obtain:

$ h \nabla^2 h + \nabla h \cdot \nabla h = 0 $

Conditions

Variables

$R_{sk}$

Hydraulic resistance of the kth layer

$kg/m^4s$

$R_{st}$

Total Resistance Hydraulic System for Serial Layers

$kg/m^4s$

Relation

Development

The equation for the height of the water column on the ground ($h$) with the time ($t$) and the flow density in more than one dimension ($\vec{j}_s$) is:

$\displaystyle\frac{\partial h}{\partial t} = - \vec{\nabla} \cdot ( h \vec{j}_s )$

in relation to

$ \vec{j}_s = - K_s \nabla h $

results after replacing and developing the derivative

$\displaystyle\frac{\partial h}{\partial t}=-\vec{\nabla}\cdot(h\vec{j}_s)= K_s\vec{\nabla}\cdot(h\nabla h)=K_s(\vec{\nabla} h\cdot\vec{\nabla} h + h \nabla^2 h)$

which, in the stationary case, reduces to

$ h \nabla^2 h + \nabla h \cdot \nabla h = 0 $

ID:(4375, 0)

Equation

>Top, >Model

In the case of the well, we can work with a polar coordinate system and assume angular symmetry, which means that the height of the water column on the ground ($h$) depends only on the radius from center of well ($r$) and satisfies the equation

$ r \displaystyle\frac{ dh^2 }{ dr } = 2 h_0 ^2\displaystyle\frac{ r_0 }{ s_0 } $

Conditions

Variables

$S_k$

Section of the kth layer

$m^2$

$S_t$

Total Section of the Parallel Layers

$m^2$

Relation

Development

As the height of the water column on the ground ($h$) satisfies

$ h \nabla^2 h + \nabla h \cdot \nabla h = 0 $

and in polar coordinates with the radius from center of well ($r$) for the case of angular symmetry, we have

$\vec{\nabla}h = \displaystyle\frac{du}{dr}\hat{r}$

and

$\nabla ^2h = \displaystyle\frac{1}{r}\displaystyle\frac{d}{dr}\left(r\displaystyle\frac{dh}{dr}\right)$

we obtain

$\displaystyle\frac{1}{r}\displaystyle\frac{d}{dr}\left(r\displaystyle\frac{dh^2}{dr}\right)=0$

or

$r\displaystyle\frac{dh^2}{dr}=C$

with $C$ as a constant. On the other hand, the equation with the hydraulic conductivity ($K_s$) and the flow density in more than one dimension ($\vec{j}_s$)

$ \vec{j}_s = - K_s \nabla h $

in polar coordinates with rotational symmetry reduces to

$j_s = - K_s \displaystyle\frac{dh}{dr}$

which at the well's surface with the water well radius ($r_0$), the characteristic length of the flow in the ground ($s_0$), the flow at a reference point ($j_{s0}$), and the reference height of the water column ($h_0$) leads to the conclusion that with

$ s_0 \equiv \displaystyle\frac{| j_{s0} |}{ K_s h_0 }$

we have

$C=r_0\displaystyle\frac{dh^2}{dr}=r_02h_0\displaystyle\frac{dh}{dr}=2r_0h_0\displaystyle\frac{|j_{s0}|}{K_sh_0}=2h_0^2\displaystyle\frac{r_0}{s_0}$

resulting in

$ r \displaystyle\frac{ dh^2 }{ dr } = 2 h_0 ^2\displaystyle\frac{ r_0 }{ s_0 } $

ID:(4430, 0)

Equation

>Top, >Model

In the case of flow towards a well, the height of the water column on the ground ($h$) as a function of the radius from center of well ($r$) with the water well radius ($r_0$), the characteristic length of the flow in the ground ($s_0$), and the reference height of the water column ($h_0$) is represented by

This equation can be analytically solved for the given conditions, with the water well radius ($r_0$) and the reference height of the water column ($h_0$) at the well's edge, as follows:

$ \displaystyle\frac{ h }{ h_0 } = \sqrt{1 + \displaystyle\frac{ 2 r_0 }{ s_0 }\ln\left(\displaystyle\frac{ r }{ r_0 }\right)} $

Conditions

Variables

$l_c$

Length and width of a clay plate

$m$

$R$

Porosity Radio

$m$

Relation

Development

The equation for the height of the water column on the ground ($h$) as a function of the radius from center of well ($r$) with the water well radius ($r_0$), the characteristic length of the flow in the ground ($s_0$), and the reference height of the water column ($h_0$) is as follows:

$ r \displaystyle\frac{ dh^2 }{ dr } = 2 h_0 ^2\displaystyle\frac{ r_0 }{ s_0 } $

This equation can be rearranged to facilitate integration as follows:

$dh^2 = 2h_0^2\displaystyle\frac{r_0}{s_0}\displaystyle\frac{dr}{r}$

Next, by integrating both sides, we obtain the height at the well's wall with the reference height of the water column ($h_0$) and the water well radius ($r_0$):

$h^2 - h_0^2 = 2h_0^2\displaystyle\frac{r_0}{s_0}\ln\left(\displaystyle\frac{r}{r_0}\right)$

Finally, by rearranging the height of the water column on the ground ($h$), we get:

$ \displaystyle\frac{ h }{ h_0 } = \sqrt{1 + \displaystyle\frac{ 2 r_0 }{ s_0 }\ln\left(\displaystyle\frac{ r }{ r_0 }\right)} $

ID:(10593, 0)

Equation

>Top, >Model

The flow density in more than one dimension ($\vec{j}_s$) is related to the hydraulic conductivity ($K_s$), and in polar coordinates with angular symmetry the radius from center of well ($r$), this results in

Therefore, with the solution for the flux density ($j_s$) and the flow at a reference point ($j_{s0}$) given the radius from center of well ($r$), the water well radius ($r_0$), and the characteristic length of the flow in the ground ($s_0$), we obtain:

$ \displaystyle\frac{ j_s }{ j_{s0} } = \displaystyle\frac{1}{\displaystyle\frac{ r }{ r_0 }\sqrt{1 + \displaystyle\frac{ 2 r_0 }{ s_0 }\ln\left(\displaystyle\frac{r}{r_0}\right)}}$

Conditions

Variables

$L$

Capillary length

$m$

$v_s$

Flow speed

$m/s$

$\Delta h$

Height or depth difference

$m$

$K_s$

Hydraulic conductivity

$m/s$

Relation

Development

With the solution for the height of the water column on the ground ($h$) and the reference height of the water column ($h_0$) given the radius from center of well ($r$), the water well radius ($r_0$), and the characteristic length of the flow in the ground ($s_0$), we obtain:

$ \displaystyle\frac{ h }{ h_0 } = \sqrt{1 + \displaystyle\frac{ 2 r_0 }{ s_0 }\ln\left(\displaystyle\frac{ r }{ r_0 }\right)} $

We can calculate the flow density in more than one dimension ($\vec{j}_s$) from the hydraulic conductivity ($K_s$) as follows:

$ \vec{j}_s = - K_s \nabla h $

And with the flux density ($j_s$) and the flow at a reference point ($j_{s0}$) using

$ s_0 \equiv \displaystyle\frac{| j_{s0} |}{ K_s h_0 }$

in this way, we obtain:

$ \displaystyle\frac{ j_s }{ j_{s0} } = \displaystyle\frac{1}{\displaystyle\frac{ r }{ r_0 }\sqrt{1 + \displaystyle\frac{ 2 r_0 }{ s_0 }\ln\left(\displaystyle\frac{r}{r_0}\right)}}$

ID:(4368, 0)