Parallel hydraulic elements

Storyboard

When hydraulic elements are connected in parallel, the flow is distributed among them, while the pressure drop is the same for all of them. The sum of the individual flows results in the total flow, and therefore, the total hydraulic resistance is equal to the inverse of the sum of the inverses of the individual hydraulic resistances. On the other hand, hydraulic conductivities are summed directly.

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Mechanisms

Iframe

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Code
Concept

Mechanisms

ID:(15726, 0)



Hydraulic resistance of elements in parallel

Concept

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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_{hk} =\displaystyle\frac{8 \eta | \Delta L_k | }{ \pi R_k ^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:

$ \Delta p = R_h J_V $



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)



Hydraulic conductance of parallel elements

Concept

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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 parallel ($R_{pt}$), along with the hydraulic resistance in a network ($R_{hk}$), in

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



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

$ R_{hk} = \displaystyle\frac{1}{ G_{hk} }$



leads to the parallel total hydraulic conductance ($G_{pt}$) can be calculated with:

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

ID:(15946, 0)



Process for the addition of hydraulic resistances in parallel

Description

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First, values for the hydraulic resistance in a network ($R_{hk}$) are calculated using the variables the viscosity ($\eta$), the cylinder k radio ($R_k$), and the tube k length ($\Delta L_k$) through the following equation:

$ R_{hk} =\displaystyle\frac{8 \eta | \Delta L_k | }{ \pi R_k ^4}$



These values are then summed to obtain the total hydraulic resistance in series ($R_{st}$):

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



With this result, it is possible to calculate the variación de la Presión ($\Delta p$) for the total hydraulic resistance in parallel ($R_{pt}$) using:



Once the variación de la Presión ($\Delta p$) is determined, the volume flow in a network ($J_{Vk}$) is calculated via:



For the case of three resistances, the calculations can be visualized in the following chart:

ID:(11070, 0)



Model

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Parameters

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$R_k$
R_k
Cylinder k radio
m
$G_{hk}$
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
$\pi$
pi
Pi
rad
$R_{pt}$
R_pt
Total hydraulic resistance in parallel
kg/m^4s
$\Delta p$
Dp
Variación de la Presión
Pa
$\eta$
eta
Viscosity
Pa s

Variables

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$J_{Vt}$
J_Vt
Flujo de Volumen Total
m^3/s
$\Delta L_k$
DL_k
Tube k length
m
$J_V$
J_V
Volume flow
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

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

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


$ \Delta p = R_h J_V $

Dp = R_h * J_V


$ \Delta p = R_h J_V $

Dp = R_h * J_V


$ G_{hk} =\displaystyle\frac{ \pi R_k ^4}{8 \eta | \Delta L_k | }$

G_h = pi * R ^4/(8* eta * abs( DL ))


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

G_pt = @SUM( G_hk , k )


$ J_{Vt} = G_{pt} \Delta p $

J_V = G_h * Dp


$ J_{Vk} = G_{hk} \Delta p $

J_V = G_h * Dp


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

J_Vt =sum_k J_Vk


$ R_{pt} = \displaystyle\frac{1}{ G_{pt} }$

R_h = 1/ G_h


$ R_{hk} = \displaystyle\frac{1}{ G_{hk} }$

R_h = 1/ G_h


$ R_{hk} =\displaystyle\frac{8 \eta | \Delta L_k | }{ \pi R_k ^4}$

R_h =8* eta * abs( DL )/( pi * R ^4)

ID:(15731, 0)



Darcy's law and hydraulic conductance (1)

Equation

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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_{Vt} = G_{pt} \Delta p $

$ J_V = G_h \Delta p $

$G_h$
$G_{pt}$
Parallel total hydraulic conductance
$m^4s/kg$
10136
$\Delta p$
Variación de la Presión
$Pa$
6673
$J_V$
$J_{Vt}$
Flujo de Volumen Total
$m^3/s$
6611

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_{hk} =\displaystyle\frac{ \pi R_k ^4}{8 \eta | \Delta L_k | }$



to arrive at:

$ J_V = G_h \Delta p $

ID:(14471, 1)



Darcy's law and hydraulic conductance (2)

Equation

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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_{Vk} = G_{hk} \Delta p $

$ J_V = G_h \Delta p $

$G_h$
$G_{hk}$
Hydraulic conductance in a network
$m^4s/kg$
10134
$\Delta p$
Variación de la Presión
$Pa$
6673
$J_V$
$J_{Vk}$
Volume flow in a network
$m^3/s$
10133

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_{hk} =\displaystyle\frac{ \pi R_k ^4}{8 \eta | \Delta L_k | }$



to arrive at:

$ J_V = G_h \Delta p $

ID:(14471, 2)



Hydraulic Conductance of a Pipe

Equation

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With the tube radius ($R$), the viscosity ($\eta$) and the tube length ($\Delta L$) we have that a hydraulic conductance ($G_h$) is:

$ G_{hk} =\displaystyle\frac{ \pi R_k ^4}{8 \eta | \Delta L_k | }$

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

$G_h$
$G_{hk}$
Hydraulic conductance in a network
$m^4s/kg$
10134
$\pi$
Pi
3.1415927
$rad$
5057
$\Delta L$
$\Delta L_k$
Tube k length
$m$
10375
$R$
$R_k$
Cylinder k radio
$m$
10376
$\eta$
Viscosity
$Pa s$
5422

ID:(15102, 0)



Sum of parallel flows

Equation

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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)



Hydraulic conductance of elements in parallel

Equation

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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

$ J_{Vk} = G_{hk} \Delta p $



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 conductance (1)

Equation

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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_{pt} = \displaystyle\frac{1}{ G_{pt} }$

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

$G_h$
$G_{pt}$
Parallel total hydraulic conductance
$m^4s/kg$
10136
$R_h$
$R_{pt}$
Total hydraulic resistance in parallel
$kg/m^4s$
5429

ID:(15092, 1)



Hydraulic conductance (2)

Equation

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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_{hk} = \displaystyle\frac{1}{ G_{hk} }$

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

$G_h$
$G_{hk}$
Hydraulic conductance in a network
$m^4s/kg$
10134
$R_h$
$R_{hk}$
Hydraulic resistance in a network
$kg/m^4s$
9887

ID:(15092, 2)



Hydraulic resistance of a tube

Equation

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Since the hydraulic resistance ($R_h$) is equal to the inverse of the hydraulic conductance ($G_h$), it can be calculated from the expression of the latter. In this way, we can identify parameters related to geometry (the tube length ($\Delta L$) and the tube radius ($R$)) and the type of liquid (the viscosity ($\eta$)), which can be collectively referred to as a hydraulic resistance ($R_h$):

$ R_{hk} =\displaystyle\frac{8 \eta | \Delta L_k | }{ \pi R_k ^4}$

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

$R_h$
$R_{hk}$
Hydraulic resistance in a network
$kg/m^4s$
9887
$\pi$
Pi
3.1415927
$rad$
5057
$\Delta L$
$\Delta L_k$
Tube k length
$m$
10375
$R$
$R_k$
Cylinder k radio
$m$
10376
$\eta$
Viscosity
$Pa s$
5422

Since the hydraulic resistance ($R_h$) is equal to the hydraulic conductance ($G_h$) as per the following equation:

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



and since the hydraulic conductance ($G_h$) is expressed in terms of the viscosity ($\eta$), the tube radius ($R$), and the tube length ($\Delta L$) as follows:

$ G_{hk} =\displaystyle\frac{ \pi R_k ^4}{8 \eta | \Delta L_k | }$



we can conclude that:

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

ID:(3629, 0)



Hydraulic resistance of parallel elements

Equation

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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

$ R_{hk} = \displaystyle\frac{1}{ G_{hk} }$



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)



Darcy's law and hydraulic resistance (1)

Equation

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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, 1)



Darcy's law and hydraulic resistance (2)

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, 2)