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Flow from a channel
Storyboard 
If there is a channel above the water table level, a flow will be created from the channel to the water table. This flow depends on the properties of the soil and the flow occurring in the channel.
ID:(2081, 0)
Flow from a channel
Concept
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.
ID:(4746, 0)
Situation that meets boundary conditions
Concept
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:
| $ h j_s = h_0 j_{s0} $ |
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.
ID:(4370, 0)
Flow height solution from a channel
Concept
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:
| $ \displaystyle\frac{ h }{ h_0 } = \sqrt{1 - \displaystyle\frac{ 2 x }{ s_0 }} $ |
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.
ID:(4374, 0)
Flux density solution from a channel
Concept
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:
| $ \displaystyle\frac{ j_s }{ j_{s0} } = \displaystyle\frac{1}{\sqrt{1 - \displaystyle\frac{ 2 x }{ s_0 }}} $ |
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.
ID:(7827, 0)
Dam I - Mina Córrego do Feijão
Concept
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.
ID:(4378, 0)
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Target : Equation To be used 
Equations
$ h \displaystyle\frac{ dh }{ dx } = -\displaystyle\frac{ h_0 ^2 }{ s_0 } $
h * @DIFF( h , x, 1) = - h_0 ^2/ s_0
$ h j_s = h_0 j_{s0} $
h * j_s = h_0 * j_{s0}
$ \displaystyle\frac{ h }{ h_0 } = \sqrt{1 - \displaystyle\frac{ 2 x }{ s_0 }} $
h / h_0 = sqrt(1 - 2* x / s_0 )
$ \displaystyle\frac{ j_s }{ j_{s0} } = \displaystyle\frac{1}{\sqrt{1 - \displaystyle\frac{ 2 x }{ s_0 }}} $
j / j_s0 = 1/sqrt(1 - 2* x / s_0 )
ID:(15820, 0)
Static solution in one dimension
Equation
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} $ |
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 $t$.
ID:(15107, 0)
Flow equation from a channel
Equation
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 }{ s_0 } $ |
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)
Flow height from a channel
Equation
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:
| $ \displaystyle\frac{ h }{ h_0 } = \sqrt{1 - \displaystyle\frac{ 2 x }{ s_0 }} $ |
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 }{ 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 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 }{ s_0 }} $ |
ID:(2214, 0)
Flow density from a channel
Equation
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 }}} $ |
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 }{ 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 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 }{ s_0 }}} $ |
ID:(4742, 0)