Flow through multiple layers

Storyboard

Once the hydraulic resistance and conductivity have been calculated, it becomes possible to model a multi-layer soil system. To achieve this, it is essential to compute the total resistance and conductivity and, after establishing the overall flow, to determine the partial flows (in the case of parallel layers) or the pressure drop in each layer (in the case of series layers).

>Model

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Mechanisms

Iframe

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Code
Concept

Mechanisms

ID:(15204, 0)



Hydraulic resistance of elements in series

Concept

>Top


In the case of a sum where the elements are connected in series, the total hydraulic resistance of the system is calculated by summing the individual resistances of each element.



One way to model a tube with varying cross-section is to divide it into sections with constant radius and then sum the hydraulic resistances in series. Suppose we have a series of the hydraulic resistance in a network ($R_{hk}$), which depends on the viscosity ($\eta$), the cylinder k radio ($R_k$), and the tube k length ($\Delta L_k$) via the following equation:

$ R_h =\displaystyle\frac{8 \eta | \Delta L | }{ \pi R ^4}$



In each segment, there will be a pressure difference in a network ($\Delta p_k$) with the hydraulic resistance in a network ($R_{hk}$) and the volume flow ($J_V$) to which Darcy's Law is applied:



the total pressure difference ($\Delta p_t$) will be equal to the sum of the individual pressure difference in a network ($\Delta p_k$):

$ \Delta p_t =\displaystyle\sum_k \Delta p_k $



therefore,

$\Delta p_t=\displaystyle\sum_k \Delta p_k=\displaystyle\sum_k (R_{hk}J_V)=\left(\displaystyle\sum_k R_{hk}\right)J_V\equiv R_{st}J_V$



Thus, the system can be modeled as a single conduit with the hydraulic resistance calculated as the sum of the individual components:

$ R_{st} =\displaystyle\sum_k R_{hk} $

ID:(3630, 0)



Hydraulic conductance of elements in series

Concept

>Top


In the case of a sum where the elements are connected in series, the total hydraulic conductance of the system is calculated by adding the individual hydraulic conductances of each element.



the total hydraulic resistance in series ($R_{st}$), along with the hydraulic resistance in a network ($R_{hk}$), in

$ R_{st} =\displaystyle\sum_k R_{hk} $



and along with the hydraulic conductance in a network ($G_{hk}$) and the equation



leads to the total Series Hydraulic Conductance ($G_{st}$) can be calculated with:

$\displaystyle\frac{1}{ G_{st} }=\displaystyle\sum_k\displaystyle\frac{1}{ G_{hk} }$



$\Delta p_k = \displaystyle\frac{J_{Vk}}{K_{hk}}$



So, the sum of the inverse of the hydraulic conductance in a network ($G_{hk}$) will be equal to the inverse of the total Series Hydraulic Conductance ($G_{st}$).

ID:(11067, 0)



Flow through serial soil layers

Concept

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A situation in the soil where the elements are connected in series occurs when water infiltrates vertically through several layers, eventually ending up in the water table. In this case, the column Section ($S$) remains constant, while each layer has a different width that acts as the width of the kth layer ($L_k$).



In this scenario, hydraulic resistances are directly summed, and their values depend on the type of soil, and therefore, on the hydraulic conductivity in the kth layer ($K_{sk}$) and the width of the kth layer ($L_k$).

ID:(936, 0)



Hydraulic resistance of elements in parallel

Concept

>Top


One efficient way to model a tube with varying cross-sections is to divide it into sections with constant radii and then sum the hydraulic resistances in series. Suppose we have a series of elements the hydraulic resistance in a network ($R_{hk}$), whose resistance depends on the viscosity ($\eta$), the cylinder k radio ($R_k$), and the tube k length ($\Delta L_k$), according to the following equation:

$ R_h =\displaystyle\frac{8 \eta | \Delta L | }{ \pi R ^4}$



In each element, we consider a pressure difference in a network ($\Delta p_k$) along with the hydraulic resistance in a network ($R_{hk}$) and the volumetric flow rate the volume flow ($J_V$), where Darcy's law is applied:



The total resistance of the system, the flujo de Volumen Total ($J_{Vt}$), is equal to the sum of the individual hydraulic resistances volume flow in a network ($J_{Vk}$) of each section:

$ J_{Vt} =\displaystyle\sum_k J_{Vk} $



Thus, we have:

$J_{Vt}=\displaystyle\sum_k \Delta J_{Vk}=\displaystyle\sum_k \displaystyle\frac{\Delta p_k}{R_{hk}}=\left(\displaystyle\sum_k \displaystyle\frac{1}{R_{hk}}\right)\Delta p\equiv \displaystyle\frac{1}{R_{pt}}J_V$



Therefore, the system can be modeled as a single conduit with a total hydraulic resistance calculated by summing the individual components:

$\displaystyle\frac{1}{ R_{pt} }=\sum_k\displaystyle\frac{1}{ R_{hk} }$

ID:(11068, 0)



Total flow (2)

Equation

>Top, >Model


The flujo de Volumen Total ($J_{Vt}$) represents the total sum of the individual contributions from the volume flow 1 ($J_{V1}$) and the volume flow 2 ($J_{V2}$), from the elements connected in parallel:

$ J_{Vt} = J_{V1} + J_{V2} $

$J_{Vt}$
Flujo de Volumen Total
$m^3/s$
6611
$J_{V1}$
Volume flow 1
$m^3/s$
8478
$J_{V2}$
Volume flow 2
$m^3/s$
8479

ID:(12800, 0)



Flow through parallel soil layers

Concept

>Top


A situation in the soil where the elements are connected in parallel occurs when water flows through different layers in parallel. If the layers have a slope, a pressure difference is generated. If the layers have a similar thickness, the pressure difference will be the same in all layers. In this case, the sample length ($\Delta L$) is constant, while each layer has a different the section of the kth layer ($S_k$).



In this situation, hydraulic conductivities are directly summed, and their values depend on the type of soil, and therefore, on the hydraulic conductivity in the kth layer ($K_{sk}$) and the section of the kth layer ($S_k$).

ID:(4373, 0)



Model

Top

>Top



Parameters

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$d$
d
Horizontal Distance
m
$G_h$
G_h
Hydraulic conductance
m^4s/kg
$G_{hk}$
G_hk
Hydraulic conductance in a network
m^4s/kg
$R_h$
R_h
Hydraulic resistance
kg/m^4s
$R_{hk}$
R_hk
Hydraulic resistance in a network
kg/m^4s
$G_{pt}$
G_pt
Parallel total hydraulic conductance
m^4s/kg
$\rho_s$
rho_s
Solid Density
kg/m^3
$R_{pt}$
R_pt
Total hydraulic resistance in parallel
kg/m^4s
$R_{st}$
R_st
Total hydraulic resistance in series
kg/m^4s
$G_{st}$
G_st
Total Series Hydraulic Conductance
m^4s/kg
$\Delta p$
Dp
Variación de la Presión
Pa

Variables

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$\rho_b$
rho_b
Dry bulk density
kg/m^3
$J_{Vt}$
J_Vt
Flujo de Volumen Total
m^3/s
$h$
h
Height increase
m
$H$
H
Layer height
m
$g_a$
g_a
Mass fraction of sand in the sample
-
$V_p$
V_p
Pore volume
m^3
$f$
f
Porosity
-
$\theta_c$
theta_c
Porosity-clay ratio factor
-
$\Delta p_k$
Dp_k
Pressure difference in a network
Pa
$S_1$
S_1
Section in point 1
m^2
$S_2$
S_2
Section in point 2
m^2
$S$
S
Section or Area
m^2
$m$
m
Slope
-
$V_c$
V_c
Solid volume of clay
m^3
$\Delta p_t$
Dp_t
Total pressure difference
Pa
$J_V$
J_V
Volume flow
m^3/s
$J_{V1}$
J_V1
Volume flow 1
m^3/s
$J_{V2}$
J_V2
Volume flow 2
m^3/s
$J_{Vk}$
J_Vk
Volume flow in a network
m^3/s

Calculations


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

Calculations

Symbol
Equation
Solved
Translated

Calculations

Symbol
Equation
Solved
Translated

Variable Given Calculate Target : Equation To be used




Equations

#
Equation

$\displaystyle\frac{1}{ G_{st} }=\displaystyle\sum_k\displaystyle\frac{1}{ G_{hk} }$

1/ G_st = @SUM( 1/ G_hk, k )


$\displaystyle\frac{1}{ R_{pt} }=\sum_k\displaystyle\frac{1}{ R_{hk} }$

1/ R_pt =@SUM( 1/ R_hk , k )


$ \Delta p = R_h J_V $

Dp = R_h * J_V


$ \Delta p_t =\displaystyle\sum_k \Delta p_k $

Dp_t =sum_k Dp_k


$ G_h = \displaystyle\frac{ K_s }{ \rho_w g }\displaystyle\frac{ S }{ \Delta L }$

G_h = K_s * S /( rho_w * g * DL )


$ G_h = \displaystyle\frac{ r_0 ^2}{8 \eta q_0 }\displaystyle\frac{f ^3}{(1- f )^2 }\displaystyle\frac{ S }{ \Delta L }$

G_h = r_0 ^2* f ^3 * S /( 8* eta * q_0 * (1- f )^2* DL )


$ G_{pt} =\displaystyle\sum_k G_{hk} $

G_pt = @SUM( G_hk , k )


$ G_{pt} = \displaystyle\sum_k\displaystyle\frac{ K_{sk} }{ \rho_w g }\displaystyle\frac{ S_k }{ L }$

G_pt = @SUM( K_sk * S_k /( rho_w * g * L ), k )


$ J_V = G_h \Delta p $

J_V = G_h * Dp


$ J_{Vt} = J_{V1} + J_{V2} $

J_Vt = J_V1 + J_V2


$ J_{Vt} =\displaystyle\sum_k J_{Vk} $

J_Vt =sum_k J_Vk


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

K_s = r_0 ^2 * f ^3 * rho_w * g /(8* q_0 *(1- f )^2* eta )


$ R_h = \displaystyle\frac{1}{ G_h }$

R_h = 1/ G_h


$ R_h = \displaystyle\frac{8 \eta q_0 }{ r_0 ^2}\displaystyle\frac{(1- f )^2 }{f ^3}\displaystyle\frac{ \Delta L }{ S }$

R_h = 8* eta * q_0 * (1- f )^2* DL /( r_0 ^2* f ^3 * S )


$ R_h = \displaystyle\frac{ \rho_w g }{ K_s }\displaystyle\frac{ \Delta L }{ S }$

R_h = rho_w * g * DL /( K_s * S )


$ R_{st} = \displaystyle\sum_k\displaystyle\frac{ \rho_w g }{ K_{sk} }\displaystyle\frac{ L_k }{ S }$

R_st = @SUM( rho_w * g * L_k /( K_sk * S ), k )


$ R_{st} =\displaystyle\sum_k R_{hk} $

R_st =@SUM( R_hk , k )

ID:(15223, 0)



Soil hydraulic conductivity

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 }$

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 ($\eta$).

ID:(4739, 0)



Soil hydraulic conductance

Equation

>Top, >Model


As the total flow ($J_{Vt}$) 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 column Section ($S$), and the sample length ($\Delta L$), it is equal to:

$ J_{Vt} =-\displaystyle\frac{ r_0 ^2}{8 \eta q_0 }\displaystyle\frac{ f ^3}{(1- f )^2}\displaystyle\frac{ S }{ \Delta L } \Delta p $



Therefore, the hydraulic conductance ($G_h$) is equal to:

$ G_h = \displaystyle\frac{ r_0 ^2}{8 \eta q_0 }\displaystyle\frac{f ^3}{(1- f )^2 }\displaystyle\frac{ S }{ \Delta L }$

ID:(15103, 0)



Hydraulic conductance

Equation

>Top, >Model


In the context of electrical resistance, there exists its inverse, known as electrical conductance. Similarly, what would be the hydraulic conductance ($G_h$) can be defined in terms of the hydraulic resistance ($R_h$) through the expression:

$ R_h = \displaystyle\frac{1}{ G_h }$

$G_h$
Hydraulic conductance
$m^4s/kg$
10124
$R_h$
Hydraulic resistance
$kg/m^4s$
5424

ID:(15092, 0)



Hydraulic resistance of a component

Equation

>Top, >Model




$ R_h = \displaystyle\frac{8 \eta q_0 }{ r_0 ^2}\displaystyle\frac{(1- f )^2 }{f ^3}\displaystyle\frac{ \Delta L }{ S }$

$S_1$
Section in point 1
$m^2$
5257
$S_2$
Section in point 2
$m^2$
6017
$S$
Section or Area
$m^2$
5405

With Darcy's law, where the pressure difference ($\Delta p$) equals the hydraulic resistance ($R_h$) and the total flow ($J_{Vt}$):

$ \Delta p = R_h J_V $



Thus, with the equation for the soil with the section Flow ($S$), the radius of a generic grain ($r_0$), the viscosity ($\eta$), the generic own porosity ($q_0$), the porosity ($f$), the pressure difference ($\Delta p$), and the sample length ($\Delta L$):

$ J_{Vt} =-\displaystyle\frac{ r_0 ^2}{8 \eta q_0 }\displaystyle\frac{ f ^3}{(1- f )^2}\displaystyle\frac{ S }{ \Delta L } \Delta p $



Therefore, the hydraulic resistance ($R_h$) is:

$ R_h = \displaystyle\frac{8 \eta q_0 }{ r_0 ^2}\displaystyle\frac{(1- f )^2 }{f ^3}\displaystyle\frac{ \Delta L }{ S }$

ID:(10594, 0)



Hydraulic resistance as a function of conductivity

Equation

>Top, >Model


Calculating the hydraulic resistance ($R_h$) with the viscosity ($\eta$), the generic own porosity ($q_0$), the radius of a generic grain ($r_0$), the porosity ($f$), the sample length ($\Delta L$), and the column Section ($S$) using

$ R_h = \displaystyle\frac{8 \eta q_0 }{ r_0 ^2}\displaystyle\frac{(1- f )^2 }{f ^3}\displaystyle\frac{ \Delta L }{ S }$



which can be rewritten using the expression for the hydraulic conductivity ($K_s$) with the liquid density ($\rho_w$) and the gravitational Acceleration ($g$), resulting in

$ R_h = \displaystyle\frac{ \rho_w g }{ K_s }\displaystyle\frac{ \Delta L }{ S }$

$\rho_b$
Dry bulk density
$kg/m^3$
5804
$f_k$
Mass fraction of sand in the sample
$-$
5797
$f$
Porosity
$-$
5805
$\theta_c$
Porosity-clay ratio factor
$-$
8247
$\rho_s$
Solid Density
$kg/m^3$
4944

Calculating the hydraulic resistance ($R_h$) using the viscosity ($\eta$), the generic own porosity ($q_0$), the radius of a generic grain ($r_0$), the porosity ($f$), the sample length ($\Delta L$), and the column Section ($S$) through

$ R_h = \displaystyle\frac{8 \eta q_0 }{ r_0 ^2}\displaystyle\frac{(1- f )^2 }{f ^3}\displaystyle\frac{ \Delta L }{ S }$



and utilizing the expression for the hydraulic conductivity ($K_s$)

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



is obtained after replacing the common factors

$ R_h = \displaystyle\frac{ \rho_w g }{ K_s }\displaystyle\frac{ \Delta L }{ S }$

ID:(10635, 0)



Conductance as a function of hydraulic conductivity

Equation

>Top, >Model


As the hydraulic resistance ($R_h$) is related to the hydraulic conductivity ($K_s$), the liquid density ($\rho_w$), the gravitational Acceleration ($g$), the column Section ($S$), and the sample length ($\Delta L$), it is expressed as

$ R_h = \displaystyle\frac{ \rho_w g }{ K_s }\displaystyle\frac{ \Delta L }{ S }$



Since the hydraulic conductance ($G_h$) is the inverse of the hydraulic resistance ($R_h$), we can conclude that

$ G_h = \displaystyle\frac{ K_s }{ \rho_w g }\displaystyle\frac{ S }{ \Delta L }$

$V_p$
Pore volume
$m^3$
5806
$\theta_c$
Porosity-clay ratio factor
$-$
8247
$V_c$
Solid volume of clay
$m^3$
6556

As the hydraulic resistance ($R_h$) is associated with the hydraulic conductivity ($K_s$), the liquid density ($\rho_w$), the gravitational Acceleration ($g$), the column Section ($S$), and the sample length ($\Delta L$), it is expressed as

$ R_h = \displaystyle\frac{ \rho_w g }{ K_s }\displaystyle\frac{ \Delta L }{ S }$



And the relationship for the hydraulic conductance ($G_h$)

$ R_h = \displaystyle\frac{1}{ G_h }$



leads to

$ G_h = \displaystyle\frac{ K_s }{ \rho_w g }\displaystyle\frac{ S }{ \Delta L }$

ID:(10592, 0)



Sum of series pressures

Equation

>Top, >Model


The total pressure difference ($\Delta p_t$) in relation to the various pressure difference in a network ($\Delta p_k$), leading us to the following conclusion:

$ \Delta p_t =\displaystyle\sum_k \Delta p_k $

$\Delta p_k$
Pressure difference in a network
$Pa$
10132
$\Delta p_t$
Total pressure difference
$Pa$
9842

ID:(4377, 0)



Darcy's law and hydraulic resistance

Equation

>Top, >Model


Darcy rewrites the Hagen Poiseuille equation so that the pressure difference ($\Delta p$) is equal to the hydraulic resistance ($R_h$) times the volume flow ($J_V$):

$ \Delta p = R_h J_V $

$R_h$
Hydraulic resistance
$kg/m^4s$
5424
$\Delta p$
Variación de la Presión
$Pa$
6673
$J_V$
Volume flow
$m^3/s$
5448

The volume flow ($J_V$) can be calculated from the hydraulic conductance ($G_h$) and the pressure difference ($\Delta p$) using the following equation:

$ J_V = G_h \Delta p $



Furthermore, using the relationship for the hydraulic resistance ($R_h$):

$ R_h = \displaystyle\frac{1}{ G_h }$



results in:

$ \Delta p = R_h J_V $

ID:(3179, 0)



Hydraulic resistance of elements in series

Equation

>Top, >Model


When there are multiple hydraulic resistances connected in series, we can calculate the total hydraulic resistance in series ($R_{st}$) by summing the hydraulic resistance in a network ($R_{hk}$), as expressed in the following formula:

$ R_{st} =\displaystyle\sum_k R_{hk} $

$R_{hk}$
Hydraulic resistance in a network
$kg/m^4s$
9887
$R_{st}$
Total hydraulic resistance in series
$kg/m^4s$
5428

One way to model a tube with varying cross-section is to divide it into sections with constant radius and then sum the hydraulic resistances in series. Suppose we have a series of the hydraulic resistance in a network ($R_{hk}$), which depends on the viscosity ($\eta$), the cylinder k radio ($R_k$), and the tube k length ($\Delta L_k$) via the following equation:

$ R_h =\displaystyle\frac{8 \eta | \Delta L | }{ \pi R ^4}$



In each segment, there will be a pressure difference in a network ($\Delta p_k$) with the hydraulic resistance in a network ($R_{hk}$) and the volume flow ($J_V$) to which Darcy's Law is applied:



the total pressure difference ($\Delta p_t$) will be equal to the sum of the individual pressure difference in a network ($\Delta p_k$):

$ \Delta p_t =\displaystyle\sum_k \Delta p_k $



therefore,

$\Delta p_t=\displaystyle\sum_k \Delta p_k=\displaystyle\sum_k (R_{hk}J_V)=\left(\displaystyle\sum_k R_{hk}\right)J_V\equiv R_{st}J_V$



Thus, the system can be modeled as a single conduit with the hydraulic resistance calculated as the sum of the individual components:

$ R_{st} =\displaystyle\sum_k R_{hk} $

ID:(3180, 0)



Hydraulic conductancia of elements in series

Equation

>Top, >Model


In the case of hydraulic resistances in series, the inverse of the total Series Hydraulic Conductance ($G_{st}$) is calculated by summing the inverses of each the hydraulic conductance in a network ($G_{hk}$):

$\displaystyle\frac{1}{ G_{st} }=\displaystyle\sum_k\displaystyle\frac{1}{ G_{hk} }$

$G_{hk}$
Hydraulic conductance in a network
$m^4s/kg$
10134
$G_{st}$
Total Series Hydraulic Conductance
$m^4s/kg$
10135

The total hydraulic resistance in series ($R_{st}$), along with the hydraulic resistance in a network ($R_{hk}$), in

$ R_{st} =\displaystyle\sum_k R_{hk} $



and along with the hydraulic conductance in a network ($G_{hk}$) and the equation



leads to the total Series Hydraulic Conductance ($G_{st}$) can be calculated with:

$\displaystyle\frac{1}{ G_{st} }=\displaystyle\sum_k\displaystyle\frac{1}{ G_{hk} }$

ID:(3633, 0)



Hydraulic resistance layers in series

Equation

>Top, >Model


Since each the hydraulic resistance of the kth layer ($R_{sk}$), which is a function of the liquid density ($\rho_w$), the gravitational Acceleration ($g$), the layer section ($S$), the width of the kth layer ($L_k$), and the hydraulic conductivity in the kth layer ($K_{sk}$), is

$ R_h = \displaystyle\frac{ \rho_w g }{ K_s }\displaystyle\frac{ \Delta L }{ S }$



it follows that the total hydraulic resistance in series ($R_{st}$) is

$ R_{st} = \displaystyle\sum_k\displaystyle\frac{ \rho_w g }{ K_{sk} }\displaystyle\frac{ L_k }{ S }$

$H$
Layer height
$m$
8239

ID:(4741, 0)



Sum of parallel flows

Equation

>Top, >Model


The sum of soil layers in parallel, denoted as the total flow ($J_{Vt}$), is equal to the sum of the volume flow in a network ($J_{Vk}$):

$ J_{Vt} =\displaystyle\sum_k J_{Vk} $

$J_{Vt}$
Flujo de Volumen Total
$m^3/s$
6611
$J_{Vk}$
Volume flow in a network
$m^3/s$
10133

.

ID:(4376, 0)



Darcy's law and hydraulic conductance

Equation

>Top, >Model


With the introduction of the hydraulic conductance ($G_h$), we can rewrite the Hagen-Poiseuille equation with the pressure difference ($\Delta p$) and the volume flow ($J_V$) using the following equation:

$ J_V = G_h \Delta p $

$G_h$
Hydraulic conductance
$m^4s/kg$
10124
$\Delta p$
Variación de la Presión
$Pa$
6673
$J_V$
Volume flow
$m^3/s$
5448

If we examine the Hagen-Poiseuille law, which allows us to calculate the volume flow ($J_V$) from the tube radius ($R$), the viscosity ($\eta$), the tube length ($\Delta L$), and the pressure difference ($\Delta p$):

$ J_V =-\displaystyle\frac{ \pi R ^4}{8 \eta }\displaystyle\frac{ \Delta p }{ \Delta L }$



we can introduce the hydraulic conductance ($G_h$), defined in terms of the tube length ($\Delta L$), the tube radius ($R$), and the viscosity ($\eta$), as follows:

$ G_h =\displaystyle\frac{ \pi R ^4}{8 \eta | \Delta L | }$



to arrive at:

$ J_V = G_h \Delta p $

ID:(14471, 0)



Hydraulic conductance of elements in parallel

Equation

>Top, >Model


The parallel total hydraulic conductance ($G_{pt}$) is calculated with the sum of the hydraulic conductance in a network ($G_{hk}$):

$ G_{pt} =\displaystyle\sum_k G_{hk} $

$G_{hk}$
Hydraulic conductance in a network
$m^4s/kg$
10134
$G_{pt}$
Parallel total hydraulic conductance
$m^4s/kg$
10136

With the total flow ($J_{Vt}$) being equal to the volume flow in a network ($J_{Vk}$):

$ J_{Vt} =\displaystyle\sum_k J_{Vk} $



and with the pressure difference ($\Delta p$) and the hydraulic conductance in a network ($G_{hk}$), along with the equation



for each element, it leads us to the conclusion that with the parallel total hydraulic conductance ($G_{pt}$),

$J_{Vt}=\displaystyle\sum_k J_{Vk} = \displaystyle\sum_k G_{hk}\Delta p = G_{pt}\Delta p$



we have

$ G_{pt} =\displaystyle\sum_k G_{hk} $

.

ID:(3634, 0)



Hydraulic resistance of parallel elements

Equation

>Top, >Model


The total hydraulic resistance in parallel ($R_{pt}$) can be calculated as the inverse of the sum of the hydraulic resistance in a network ($R_{hk}$):

$\displaystyle\frac{1}{ R_{pt} }=\sum_k\displaystyle\frac{1}{ R_{hk} }$

$R_{hk}$
Hydraulic resistance in a network
$kg/m^4s$
9887
$R_{pt}$
Total hydraulic resistance in parallel
$kg/m^4s$
5429

The parallel total hydraulic conductance ($G_{pt}$) combined with the hydraulic conductance in a network ($G_{hk}$) in

$ G_{pt} =\displaystyle\sum_k G_{hk} $



and along with the hydraulic resistance in a network ($R_{hk}$) and the equation



leads to the total hydraulic resistance in parallel ($R_{pt}$) via

$\displaystyle\frac{1}{ R_{pt} }=\sum_k\displaystyle\frac{1}{ R_{hk} }$

ID:(3181, 0)



Parallel Layer Hydraulic Conductance

Equation

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Considering that each the hydraulic conductance in a network ($G_{hk}$), dependent on the liquid density ($\rho_w$), the gravitational Acceleration ($g$), the soil layer length ($L$), the section of the kth layer ($S_k$), and the hydraulic conductivity in the kth layer ($K_{sk}$), is equal to:

$ G_h = \displaystyle\frac{ K_s }{ \rho_w g }\displaystyle\frac{ S }{ \Delta L }$



Therefore, the parallel total hydraulic conductance ($G_{pt}$) is calculated as:

$ G_{pt} = \displaystyle\sum_k\displaystyle\frac{ K_{sk} }{ \rho_w g }\displaystyle\frac{ S_k }{ L }$

$h$
Height increase
$m$
8243
$d$
Horizontal Distance
$m$
5114
$m$
Slope
$-$
8245

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